9702 PHYSICS MARK SCHEME for the October/November 2007 question paper

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
9702 PHYSICS
9702/02
Paper 2 (AS Structured Questions), maximum raw mark 60
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before marking began.
All Examiners are instructed that alternative correct answers and unexpected approaches in
candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills
demonstrated.
Mark schemes must be read in conjunction with the question papers and the report on the
examination.
•
CIE will not enter into discussions or correspondence in connection with these mark schemes.
CIE is publishing the mark schemes for the October/November 2007 question papers for most IGCSE,
GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level
syllabuses.
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MARK SCHEME for the October/November 2007 question paper
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GCE Advanced Subsidiary Level and GCE Advanced Level
Page 2
1
2
Mark Scheme
GCE A/AS LEVEL – October/November 2007
Syllabus
9702
(a) systematic: e.g. constant error (in all readings)
cannot be eliminated by averaging
error in measuring instrument
random: e.g.
readings scattered (equally) about true value
error due to observer
can be eliminated by averaging
(only if averaging not included for systematic)
B1
[2]
(b) 15 = π × R2 × 20
R = 0.4886 cm (accept any number of s.f.)
% uncertainty in V = 3.3 % (or 0.5/15)
% uncertainty in L = 0.5 % (or 0.1/20)
% uncertainty in R = 1.9 % (i.e. one half of the sum)
R = 0.489 ± 0.009 cm
C1
C1
C1
C1
A1
[5]
(a) 3.5 T
B1
[1]
(b) (i) distance = average speed × time (however expressed)
= 14 m
C1
A1
[2]
A1
[1]
(c) 3.5T = 14 + 5.6(T – 5)
T = 6.7 s
C1
A1
[2]
(d) (i) acceleration = (5.6 / 5 =) 1.12 m s–2
force = ma
= 75 N
C1
C1
A1
[3]
C1
A1
[2]
B1
[1]
B1
B1
[2]
C1
A1
[2]
C1
A1
[2]
(ii) distance = 5.6 × (T – 5) (or 3.5T – 14)
(ii) power = (force × speed =) {75 + 23} × 4.5
= 440 W
(allow 1/2 for 234 W, 0/2 for 338 W or 104 W)
3
Paper
02
(a) (i) potential energy: stored energy available to do work
(ii) gravitational: due to height/position of mass OR distance from mass
OR moving mass from one point to another
elastic:
due to deformation/stretching/compressing
(b) (i) height raised = (61 – {61 cos18} =) 3.0 cm
energy = (mgh = 0.051 × 9.8 × 0.030 =) 1.5 × 10–2 J
(ii) moment = force × perpendicular distance
= 0.051 × 9.8 × 0.61 × sin18
= 0.094 N m
© UCLES 2007
B1
Page 3
4
5
Mark Scheme
GCE A/AS LEVEL – October/November 2007
Paper
02
(a) brittle
B1
(b) Young modulus = stress / strain
= (9.5 × 108) / 0.013
= 7.3 × 1010 Pa (allow ± 0.1 × 1010 Pa)
C1
(c) stress = force / area
(minimum) area = (1.9 × 103) / (9.5 × 108)
= 2.0 × 10–6 m2
(max) area of cross-section = (3.2 – 2.0) × 10–6
= 1.2 × 10–6 m2
C1
C1
A1
[3]
(d) when bent, ‘top’ and ‘bottom’ edges have different extensions
with thick rod, difference is greater (than with a thin rod)
so breaks with less bending
M1
A1
A0
[2]
(a) amplitude between 6.5 squares and 7.5 squares on 3 peaks
(allow 1 mark if outside this range but between 6.0 and 8.0 squares)
correct phase (ignore lead/lag, look at x-axis only and allow ±½ square
B2
B1
[3]
(b) λ = ax / D
540 × 10–9 = (0.700 × 10–3 x) / 2.75
x = 2.12 mm
C1
C1
A1
[3]
(c) (i) same separation
bright areas brighter (1)
dark areas, no change (1)
(allow ‘contrast greater’ for 1 mark if dark/light areas not discussed)
fewer fringes observed (1) any two, 1 each
B1
A1
[1]
[2]
B2
[3]
B1
B1
[2]
(a) power = VI
current = 10.5 × 103 / 230
= 45.7 A
C1
M1
A0
[2]
(b) (i) p.d. across cable = 5.0 V
R = 5.0 / 46
= 0.11 Ω
C1
C1
A1
[3]
C1
C1
A1
[3]
(ii) smaller separation of fringes
no change in brightness
6
Syllabus
9702
(ii) R = ρL / A
0.11 = (1.8 × 10–8 × 16 × 2) / A
A = 5.3 × 10–6 m2
(wires in parallel, not series, allow max 1/3 marks)
© UCLES 2007
Page 4
Mark Scheme
GCE A/AS LEVEL – October/November 2007
(c) (i) either power = V 2 / R or power ∝ V 2
ratio = (210 / 230)2 = 0.83
(ii) resistance of cable is greater
greater power loss/fire hazard/insulation may melt
wire may melt/cable gets hot
7
Syllabus
9702
Paper
02
C1
A1
[2]
M1
A1
[2]
(a) most α-particles deviated through small angles
(accept ‘undeviated’)
few α-particles deviated through angles greater than 90°
B1
B1
[2]
(b) (i) allow 10–9 m → 10–11 m
B1
[1]
B1
[1]
(ii) allow 10–13 m → 10–15 m
(if (i) and (ii) out of range but (ii) = 10–4(i), then allow 1 mark)
(if no units or wrong units but (ii) = 10–4(i), then allow 1 mark)
© UCLES 2007
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